I'm trying to understand how sbrk works.
Here is my little code:
int main()
{
printf("end of the break : %p\n", sbrk(0));
printf("end of the break : %p\n", sbrk(10));
printf("new end of the break : %p\n\n", sbrk(0));
}
This outputs:
end of break : 0xaa6000 end of break : 0xac7000 new end of the break : 0xac700a
Why is the difference between the first 2 addresses 0xac7000 - 0xaa6000 = 21000 and not 10?
sbrk(n) increments the break by n and returns the old value of the break.
Thus:
printf("end of the break : %p\n", sbrk(0));
Output: end of the break : 0xaa6000
Initially, the break is 0xaa6000 and the sbrk call doesn't change it.
printf("end of the break : %p\n", sbrk(10));
Output: end of the break : 0xac7000
This is the value you're asking about. Above I said sbrk(0) wouldn't change the break, so why do we get a different value here?
The only thing that's happened in between the two sbrk call is the call to the first printf. Presumably the internals of your stdio implementation use malloc (e.g. to create buffers), which in turn calls sbrk itself. In other words, printf calls malloc internally, which reserves memory using sbrk.
printf("new end of the break : %p\n\n", sbrk(0));
Output: new end of the break : 0xac700a
This time we see an increment of 0xa, which matches your previous sbrk(10) call exactly. Apparently this time printf didn't need to allocate dynamic memory (or if it did, malloc was able to do everything within the space it got from the first sbrk, so it didn't have to request more from the OS).
